Sin 2x - sin x=0
Using the trigonometric identity: sin 2x=2 sinx cosx
2 sinx cosx - sinx =0
Common factor sinx
sinx ( 2 cosx -1)=0
Two options:
1) sinx=0
on the interval [0,2π), the sinx=0 for x=0 and x=π=3.1416→x=3.14

2) 2 cosx - 1=0
Solving for cosx
2 cosx-1+1=0+1
2 cosx = 1
Dividing by 2 both sides of the equation:
(2 cosx)/2=1/2
cosx=1/2

cosx is positive in first and fourth quadrant:
First quadrant cosx=1/2→x=cos^(-1) (1/2)→x=π/3=3.1416/3→x=1.05
Fourth quadrant: x=2π-π/3=(6π-π)/3→x=5π/3=5(3.1416)/3→x=5.24

Answer: Solutions: x=0, 1.05, 3.14, and 5.24

6 0

3 years ago

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57=a^4 convert this into logarithmic form

PIT_PIT [208]

I think it’s logA(57)=4

7 0

4 years ago

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Please help me with the problem

mixas84 [53]

Let's see what we're working on here

$\frac{1}{2} - 2(m) =?$
? = 0
$Simplify this →$ $\frac{1}{2}$

$\frac{1}2y} - 2(m) = ?$

$\frac{2(m)(2)}{2}$

$\frac{1 - 4(m)}{2} = ?$

$\frac{1 - 4(m)}{2} = 2$

$-4(m) + 1 = 0$
Subtract ( - )the number 1 from each of your sides on this part

4(m) = 1
Multiply( ×) the number -1 to your sides for this part of the equation

Therefore, the value of m is $\frac{1}{4}$
$m = \frac{1}{4}$

3 0

4 years ago

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Complete the square to transform the quadratic equation into the form

masha68 [24]

D is the answer :) your welcom

6 0

3 years ago